如图所示,点E,D分别在△ABC的边AB,BC上,CE和AD交于点F,若S△ABC=1,S△BDE=S△DCE=S△ACE,则S△AED等于
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![如图所示,点E,D分别在△ABC的边AB,BC上,CE和AD交于点F,若S△ABC=1,S△BDE=S△DCE=S△ACE,则S△AED等于](/uploads/image/z/1604113-25-3.jpg?t=%E5%A6%82%E5%9B%BE%E6%89%80%E7%A4%BA%2C%E7%82%B9E%2CD%E5%88%86%E5%88%AB%E5%9C%A8%E2%96%B3ABC%E7%9A%84%E8%BE%B9AB%2CBC%E4%B8%8A%2CCE%E5%92%8CAD%E4%BA%A4%E4%BA%8E%E7%82%B9F%2C%E8%8B%A5S%E2%96%B3ABC%3D1%2CS%E2%96%B3BDE%3DS%E2%96%B3DCE%3DS%E2%96%B3ACE%2C%E5%88%99S%E2%96%B3AED%E7%AD%89%E4%BA%8E)
如图所示,点E,D分别在△ABC的边AB,BC上,CE和AD交于点F,若S△ABC=1,S△BDE=S△DCE=S△ACE,则S△AED等于
如图所示,点E,D分别在△ABC的边AB,BC上,CE和AD交于点F,若S△ABC=1,S△BDE=S△DCE=S△ACE,则S△AED等于
如图所示,点E,D分别在△ABC的边AB,BC上,CE和AD交于点F,若S△ABC=1,S△BDE=S△DCE=S△ACE,则S△AED等于
S△BDE=S△DCE推得BD=DC
S△BDE=S△DCE=S△ACE推得S△ACE=S△BEC/2推得AE=AB/3
所以S△AED=S△ABD/3=(S△ABC/2)/3=S△ABC/6=1/6
因为S△BDE=S△DCE,所以D为BC中点;设D到AB边距离为h,则C到AB边距离为2h;
有1/2×AE×2h=1/2×h×BE,故BE=2AE;S△AED=1/2×h×AE,S△ABC=1/2×2h×AB=1;
AB=3AE;故S△=1/6S△ABC=1/6
AED=ACDE-ADC
S△BDE = S△DCE推得BD = DC
S△BDE = S△大商所= S△ACE按S△ACE = S△BEC / 2推AE = AB / 3
△AED = S△ABD / 3 =(S△ABC / 2)/ 3 = S△ABC / 6 = 1/6